Answer
See below
Work Step by Step
Given:
$y''-y=12e^{2t}$
and $y(0)=1\\
y'(0)=1$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(x)-sy(0)-y'(0)]-Y(s)=\frac{12}{s-2}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2-1)-(s+1)=\frac{12}{s-2}$
That is,
$Y(s)(s^2-1)=\frac{12}{s-2}+s+1=\frac{s^2-s+10}{s-2}$
Thus, we have solved for the Laplace transform of y(t). To find y itself, we must take the inverse Laplace transform. We first decompose the right-hand side into partial fractions to obtain:
$Y(s)=\frac{s^2-s+10}{(s-2)(s^2-1)}=\frac{s^2-s+10}{(s-2)(s-1)(s+1)}=\frac{2}{s+1}-\frac{5}{s-1}+\frac{4}{s-2}$
We recognize the terms on the right-hand side as being the Laplace transform of appropriate exponential functions. Taking the inverse Laplace transform yields:
$y(t)=2e^{-t}-5e^t+4e^{2t}$
